Calculus Examples

$\frac{d y}{d x} = y x - x + 4 y - 4$

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor out the greatest common factor from each group.

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Step 1.1.1.1

Group the first two terms and the last two terms.

$\frac{d y}{d x} = (y x - x) + 4 y - 4$

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d y}{d x} = x (y - 1) + 4 (y - 1)$

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $y - 1$ .

$\frac{d y}{d x} = (y - 1) (x + 4)$

Step 1.2

Multiply both sides by $\frac{1}{y - 1}$ .

$\frac{1}{y - 1} \frac{d y}{d x} = \frac{1}{y - 1} ((y - 1) (x + 4))$

Step 1.3

Cancel the common factor of $y - 1$ .

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Step 1.3.1

Cancel the common factor.

$\frac{1}{y - 1} \frac{d y}{d x} = \frac{1}{y - 1} ((y - 1) (x + 4))$

Step 1.3.2

Rewrite the expression.

$\frac{1}{y - 1} \frac{d y}{d x} = x + 4$

Step 1.4

Rewrite the equation.

$\frac{1}{y - 1} d y = (x + 4) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y - 1} d y = \int x + 4 d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y - 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y - 1$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y - 1$ .

$\frac{d}{d y} [y - 1]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y - 1$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 2.2.1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d y} [- 1]$

Step 2.2.1.1.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- 1$ with respect to $y$ is $0$ .

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u} d u = \int x + 4 d x$

Step 2.2.2

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| u |) + C_{1} = \int x + 4 d x$

Step 2.2.3

Replace all occurrences of $u$ with $y - 1$ .

$\ln (| y - 1 |) + C_{1} = \int x + 4 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\ln (| y - 1 |) + C_{1} = \int x d x + \int 4 d x$

Step 2.3.2

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$\ln (| y - 1 |) + C_{1} = \frac{1}{2} x^{2} + C_{2} + \int 4 d x$

Step 2.3.3

Apply the constant rule.

$\ln (| y - 1 |) + C_{1} = \frac{1}{2} x^{2} + C_{2} + 4 x + C_{3}$

Step 2.3.4

Simplify.

$\ln (| y - 1 |) + C_{1} = \frac{1}{2} x^{2} + 4 x + C_{4}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$\ln (| y - 1 |) = \frac{1}{2} x^{2} + 4 x + C$

Step 3

Solve for $y$ .

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Step 3.1

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (| y - 1 |)} = e^{\frac{1}{2} x^{2} + 4 x + C}$

Step 3.2

Rewrite $\ln (| y - 1 |) = \frac{1}{2} x^{2} + 4 x + C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{1}{2} x^{2} + 4 x + C} = | y - 1 |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y - 1 | = e^{\frac{1}{2} x^{2} + 4 x + C}$ .

$| y - 1 | = e^{\frac{1}{2} x^{2} + 4 x + C}$

Step 3.3.2

Combine $\frac{1}{2}$ and $x^{2}$ .

$| y - 1 | = e^{\frac{x^{2}}{2} + 4 x + C}$

Step 3.3.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y - 1 = \pm e^{\frac{x^{2}}{2} + 4 x + C}$

Step 3.3.4

Add $1$ to both sides of the equation.

$y = \pm e^{\frac{x^{2}}{2} + 4 x + C} + 1$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{\frac{x^{2}}{2} + 4 x + C}$ as $e^{\frac{x^{2}}{2} + 4 x} e^{C}$ .

$y = \pm e^{\frac{x^{2}}{2} + 4 x} e^{C} + 1$

Step 4.2

Reorder $e^{\frac{x^{2}}{2} + 4 x}$ and $e^{C}$ .

$y = \pm e^{C} e^{\frac{x^{2}}{2} + 4 x} + 1$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{\frac{x^{2}}{2} + 4 x} + 1$